Question

Let $\mathit{R}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{S}$ be rings.

1. Prove that$\mathit{f}\mathbf{:}\mathit{R}\mathbf{\times }\mathit{S}\mathbf{\to }\mathit{R}$ given by $\mathit{f}\left(\left(r,s\right)\right)\mathbf{=}\mathit{r}$is a surjective homomorphism.
2. Prove that $\mathit{g}\mathbf{:}\mathit{R}\mathbf{\times }\mathit{S}\mathbf{\to }\mathit{S}$ given by $\mathit{g}\left(\left(r,s\right)\right)\mathbf{=}\mathit{s}$ is a surjective homomorphism.
3. If both $\mathit{R}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{S}$are non-zero rings, prove that the homomorphism $\mathit{f}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{g}$are not injective.

Short answer

Hence proved:

(a) $f$is a surjective.

(b) $g$is a surjective homomorphism.

(c) $fandg$ are not injective homomorphism.

Step-by-step solution

Homomorphism

If any ring$R$has elements such that,$a,b\in R$, then, the condition for homomorphism of the function of its elements is given by:

$$\begin{gathered} f\left(a+b\right)=f\left(a\right)+f\left(b\right) \\ f\left(ab\right)=f\left(a\right)f\left(b\right) \end{gathered}$$

Solution to part (a) 

We have a function of rings as: $f:R\times S\to R$and$f\left(\left(r,s\right)\right)=r$

Clearly, the given function is surjective, then, in this case for some elements, $r_1,r_2\in Rand{s}_1,s_2\in S$, we get:

$$\begin{gathered} f\left(\left(r_1,s_1\right)+\left(r_2,s_2\right)\right)=f\left(\left(r_1+r_2\right),\left(s_1+s_2\right)\right)=f\left(\left(r_1,s_1\right)\right)+f\left(\left(r_2,s_2\right)\right) \\ f\left(\left(r_1,s_1\right)\left(r_2,s_2\right)\right)=f\left(\left(r_1{r}_2\right),\left(s_1{s}_2\right)\right)=f\left(\left(r_1,s_1\right)\right)f\left(\left(r_2,s_2\right)\right) \end{gathered}$$

Hence proved, $f$is a surjective homomorphism.

Solution to part (b) 

We have a function of rings as: $g:R\times S\to Randg\left(\left(r,s\right)\right)=s$

Clearly, the given function is surjective, then, in this case for some elements, $r_1,r_2\in Rand{s}_1,s_2\in S$, we get:

$$\begin{gathered} g\left(\left(r_1,s_1\right)+\left(r_2,s_2\right)\right)=g\left(\left(r_1+r_2\right),\left(s_1+s_2\right)\right)=g\left(\left(r_1,s_1\right)\right)+g\left(\left(r_2,s_2\right)\right) \\ g\left(\left(r_1,s_1\right)\left(r_2,s_2\right)\right)=g\left(\left(r_1{r}_2\right),\left(s_1{s}_2\right)\right)=g\left(\left(r_1,s_1\right)\right)g\left(\left(r_2,s_2\right)\right) \end{gathered}$$

Hence proved, $g$ is a surjective homomorphism.

Solution to part (c)

We have both $RandS$ as non-zero rings. So, we have:

$r\ne 0_{R}ands\ne 0_S$

In this case, we get:

$$\begin{gathered} f\left(r,0_s\right)=r=f\left(r,s\right) \\ g\left(0_r,s\right)=s=g\left(r,s\right) \end{gathered}$$

Hence proved, $fandg$ are not injective homomorphism.